[Leetcode/파이썬] 71. Simplify Path

Simplify Path

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You are given an absolute path for a Unix-style file system, which always begins with a slash '/'. Your task is to transform this absolute path into its simplified canonical path.

The rules of a Unix-style file system are as follows:

• A single period '.' represents the current directory.
• A double period '..' represents the previous/parent directory.
• Multiple consecutive slashes such as '//' and '///' are treated as a single slash '/'.
• Any sequence of periods that does not match the rules above should be treated as a valid directory or file name. For example, '...' and '....' are valid directory or file names.

The simplified canonical path should follow these rules:

• The path must start with a single slash '/'.
• Directories within the path must be separated by exactly one slash '/'.
• The path must not end with a slash '/', unless it is the root directory.
• The path must not have any single or double periods ('.' and '..') used to denote current or parent directories.

Return the simplified canonical path.

 

Example 1:

Input: path = "/home/"

Output: "/home"

Explanation:

The trailing slash should be removed.

Example 2:

Input: path = "/home//foo/"

Output: "/home/foo"

Explanation:

Multiple consecutive slashes are replaced by a single one.

Example 3:

Input: path = "/home/user/Documents/../Pictures"

Output: "/home/user/Pictures"

Explanation:

A double period ".." refers to the directory up a level (the parent directory).

Example 4:

Input: path = "/../"

Output: "/"

Explanation:

Going one level up from the root directory is not possible.

Example 5:

Input: path = "/.../a/../b/c/../d/./"

Output: "/.../b/d"

Explanation:

"..." is a valid name for a directory in this problem.

 

Constraints:

• 1 <= path.length <= 3000
• path consists of English letters, digits, period '.', slash '/' or '_'.
• path is a valid absolute Unix path.

 

class Solution:
    def simplifyPath(self, path: str) -> str:
        path_sp = path.split("/")
        stack = []

        for p in path_sp :
            if p == "" or p == ".":
                continue

            if p == ".." :
                if stack :
                    stack.pop()
                continue
            
            stack.append(p)

        return "/" + "/".join(stack)

자료구조 스택을 활용한 풀이. 문제이다. 처음에 무슨 소린가 했는데 자세히 살펴보니 스택의 append, pop 을 활용하는 문제이니 잘 보고 문제를 해결하였으면 좋겠다.